A) for \[-1<n<0\]
B) for\[n>-1\]or\[n<0\]
C) for all\[n\ne 0,-1\]
D) for no value of n
Correct Answer: B
Solution :
Given, \[2\left[ \frac{1}{2n+1}+\frac{1}{3{{(2n+1)}^{3}}}+.... \right]=\log \left( \frac{n+1}{n} \right)\] \[-1<\frac{1}{2n+1}<1\] Now, \[-1<\frac{1}{2n+1}\]or\[\frac{1}{2n+1}<1\] \[\Rightarrow \] \[-(2n+1)<1\]or\[1<2n+1\] \[\Rightarrow \] \[-2n<2\]or\[0<2n\] \[\Rightarrow \] \[-n<1\]or\[0<n\] \[\Rightarrow \] \[n>-1\]or\[0<n\]
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